Simplify and expand the following expression: $ \dfrac{3}{2k - 10}+ \dfrac{5}{2k - 18}- \dfrac{3}{k^2 - 14k + 45} $
Answer: First find a common denominator by finding the least common multiple of the denominators. Try factoring the denominators. We can factor a $2$ out of denominator in the first term: $ \dfrac{3}{2k - 10} = \dfrac{3}{2(k - 5)}$ We can factor a $2$ out of denominator in the second term: $ \dfrac{5}{2k - 18} = \dfrac{5}{2(k - 9)}$ We can factor the quadratic in the third term: $ \dfrac{3}{k^2 - 14k + 45} = \dfrac{3}{(k - 5)(k - 9)}$ Now we have: $ \dfrac{3}{2(k - 5)}+ \dfrac{5}{2(k - 9)}- \dfrac{3}{(k - 5)(k - 9)} $ The least common multiple of the denominators is: $ 4(k - 5)(k - 9)$ In order to get the first term over $4(k - 5)(k - 9)$ , multiply by $\dfrac{2(k - 9)}{2(k - 9)}$ $ \dfrac{3}{2(k - 5)} \times \dfrac{2(k - 9)}{2(k - 9)} = \dfrac{6(k - 9)}{4(k - 5)(k - 9)} $ In order to get the second term over $4(k - 5)(k - 9)$ , multiply by $\dfrac{2(k - 5)}{2(k - 5)}$ $ \dfrac{5}{2(k - 9)} \times \dfrac{2(k - 5)}{2(k - 5)} = \dfrac{10(k - 5)}{4(k - 5)(k - 9)} $ In order to get the third term over $4(k - 5)(k - 9)$ , multiply by $\dfrac{4}{4}$ $ \dfrac{3}{(k - 5)(k - 9)} \times \dfrac{4}{4} = \dfrac{12}{4(k - 5)(k - 9)} $ Now we have: $ \dfrac{6(k - 9)}{4(k - 5)(k - 9)} + \dfrac{10(k - 5)}{4(k - 5)(k - 9)} - \dfrac{12}{4(k - 5)(k - 9)} $ $ = \dfrac{ 6(k - 9) + 10(k - 5) - 12} {4(k - 5)(k - 9)} $ Expand: $ = \dfrac{6k - 54 + 10k - 50 - 12}{4k^2 - 56k + 180} $ $ = \dfrac{16k - 116}{4k^2 - 56k + 180}$ Simplify: $ = \dfrac{4k - 29}{k^2 - 14k + 45}$